Good morning bobbym,

How are you today?  Or is it still yesterday where you are?

A diagram for this problem would be great.

Unfortunately, my eyesight isn't that good.  Any chance you could make it a bit bigger?

Thanks,

Bob

hi bobbym,

Arhh, that's better.  I can see it now.

Now I just need a brain transplant, and I might understand what you're doing here. 

Bob

Always another way - The Keymaker

Since this is a thread devoted to computer math programs let's see how mathematica can be used to solve the problem experimentally.

Supposing we treat this as a discrete problem rather than a continuous one. It we examine what happens at discrete intervals between 0 and 1 and then subdivide those intervals we could view the process as a limit as the intervals approach zero.

If we start with the set {0,1/3,2/3,1} we see that there is no way to pick 3 from it that the minimum is less than 1/3. The same holds for increments of 1,1/2,1/4 and 1/5.

The first solutions occur when we start with the set {0,1/6,2/6,3/6,4/6,5/6,1}. Using combinatorics or direct count we see that there are 5 solutions, they are:

We can form a table of n subdivisions and solutions:

The table means that if we have the set, {0,1/20,2/20,3/20,...1} there are 245 ways.

A finite set with 21 elements representing points on the number line is a poor substitute for the unit interval with its infinite number of points. We need a relationship between the solutions and n. We ask mathematica to find one. It comes up with the following difference equation:

We run the recurrence in the forward direction to get n = 2 000 000.

The number of ways to pick three from the 2 000 001 points is 345677975308901235.

The probability is:

This only disagrees with 7 / 27  by less than .00000077, so we are on the right track.

Can we do better? We could, if we could solve that recurrence but it will not even fit on a piece of paper.
We need a trick.

Let's study the solutions again starting at 6.

It is well behaved at 7 and 8 and then takes a big jump at 9. Well behaved again at 10 and 11 and then another big jump. This disproportionate jump occurs at every third one. Let's see if we can understand it.

The first tool to understanding any sequence is a difference table. Old timers like Newton,Leibniz, Gauss, Euler, Legendre and Lagrange lived off of them. They are much less used and understood today. Let's form a difference table of every third one.

That is what we want a last line of constant differences. That means there is a
polynomial fit for every third entry in the table. Now we find that polynomial.

So you see hiding inside the major sequence was another smaller pattern. One much easier to deal with. This is sometimes the case with recurrences and you should look for it.

Now to finish up. The more points we use in the interval 0 to 1, the closer we approximate the continuous unit interval. We will use the calculus to get an infinite number of them.

The probability of drawing three numbers from the unit interval with the maximum - minimum < 1/3 is:

And we are done!

Thanks gAr, I appreciate you saying that. You will understand it or quickly find the hole in it.
One way or another will be good.

Sorry, there was a big typo right on the end that I fixed up. I have pi on the brain.

Okay, see you tomorrow.